But the matrix A has the same form as C, namely it has nonzero variables in a block submatrix only. On 21/10/15 14:26, Lawrence Mitchell wrote:
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On 21/10/15 14:14, Anna Kalogirou wrote:
Dear Lawrence,
I believe the code (attached) sets up the matrices/vectrors/operators correctly. However, I have one question: Due to the use of a heavyside function, the operator B on the LHS is nonzero in a block NxN, say, in the bottom right corner. For that reason, when I solved the linear system in Matlab I only considered these nonzero values in the block, because otherwise the LHS would not be invertible. How come and it doesn't complain here? So it's clear that the matrix C = Q1*Q2^T is not invertible, however, presumably you're solving for:
B = A + C
where A is full rank and invertible. So I don't immediately see why there would be a problem in trying to invert B which is what the solver does.
Lawrence
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-- Dr Anna Kalogirou Research Fellow School of Mathematics University of Leeds http://www1.maths.leeds.ac.uk/~matak/